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I need to know how to calculate 10^ a decimal power, like 10^-7.4, without a calculator, in as simple a way as possible, since I will be doing questions which only allow me about a minute to a minute and a half each. Does anyone know a good technique for this?

Edit: I'm not allowed to use any other resources besides a pencil/paper/my head.

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    $\begingroup$ With a log table? $\endgroup$ – user296602 Jan 14 '16 at 22:22
  • $\begingroup$ Not even a slide rule? $\endgroup$ – Matt Samuel Jan 16 '16 at 0:53
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Notice that $$10^{-7.4} = 10^{-7} \cdot 10^{-0.4}$$

Now take a logarithm of the second part:

$$\ln 10^{-.4} = -0.4 \ln 10 \approx -0.4 \cdot 2.3 = -0.92$$ after memorizing what the log of 10 is. Then exponentiate:

\begin{align*} 10^{-.4} = e^{-.92} &= \sum_{k = 0}^{\infty} \frac{(-.92)^k}{k!} \\ &\approx 1 - .92 + \frac{.92^2}{2} - \frac{.92^3}{6} + \frac{.92^4}{24} \end{align*}

correct to at least two decimals: This is an alternating series, and so the error can be estimated by the next term, which is substantially less than $1/120$. Calculating this by hand should only take little bit of time, and I get about $0.40$ (although I did use a calculator).

So my answer would be $4.0 \times 10^{-8}$, which compares well to the more precise answer of $3.98107 \times 10^{-8}$ from a calculator. Had I computed one more term in the series expansion, I would have gotten $3.978 \times 10^{-8}$, which is quite good.


Alternative solution: Memorize that $10^{0.1} \approx 1.259$ and go from there.

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Notice that $$10^{-7.4} = 10^{-8} \cdot 10^{0.6}$$

Now calc the second part:

$$10^{0.1} = 1.2589 $$ $$10^{0.6} = 1.2589^6$$

so the answer is $$10^{-8} \cdot 3.9806 $$

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